Question

Griffiths Introductory to Quantum Mechanics (3rd Edition): Problem 7.9

Problem 7.9 Consider a particle of mass m that is free to move in a one- dimensional region of length L that closes on itself (for instance, a bea that slides frictionlessly on a circular wire of circumference L, as Problem 2.46) (a) Show that the stationary states can be written in the form 2π inx/L where n 0, t l, 2, , and the allowed energies are In Notice that-with the exception of the ground state (n = 0)-these are all doubly degenerate. (b) Now suppose we introduce the perturbatiorn where a 《 L. (This puts a little "dimple" in the potential at x = 0, as though we bent the wire slightly to make a "trap".) Find the first-order correction to En, using Equation 33. Hint: To evaluate the integrals, exploit the fact that a«L to extend the limits from L/2 too0; after all, H' is essentially zero outside (c) What are the "good" linear combinations of Vn and ψ-, for this problem? (Hint: use Eq. 7.27) Show that with these states you get the first-order correction using Equation 7-9. (d) Find a hermitian operator A that fits the requirements of the theorem, and show that the simultaneous eigenstates of HO and A are precisely the ones you used in (c).

Answer 1

   The problem of a particle confined a one-dimensional ring is similar to the particle in a 1D box. In contrast to the particle in a box, the eigenfunctions are linearly independent. Therefore all eigenvalues, (except m=0) are doubly degenerate. Look at the detailed solution and after that watch carefully how perturbation theory works in presence of degeneracy.
schrodinger equation: - h^2/2M d^2 psi/dx^2 = E psi implies d^2 psi/dx^2 + k^2 psi = 0, k = squareroot 2 M E/h^2 so, psi (x) = Ae^ikx + Be^- ikx Applying Boundary condition psi (x + L = psi (x) at x = 0 gives, Ae^ikx e^ikL + Be^- ikx e^- ikL = Ae^ikx + Be^- ikx implies Ae^ikL + Be^- ikL = A + B (putting x = 0) For, x = pi/2k, Ae^ikl - Be^- ikl = A - B from (1) and (2) 2Ae^ikl = 2A either A = 0 on e^ikl = 1 Be^- ikl = B kl = 2 m pi (m = 0, plusminus 1, plusminus 2, .....) so, for every positive m there are two solution: {psi^+_m = Ae^i 2m pi x/L psi^-_m = B e^- i 2 m pi x/L for n = 0, only one solution exist